Remarks Every set of invariants is realized by a formal Legendrian knot, with the additional note in case (a) that the parity of \(tb(L)\)  is determined by \(r(L)\) . However note that Proposition  REF  is false if \(n=1\) : there is no Legendrian realizing a formal Legendrian unknot with \(tb = 0\) . For \(n>3\)  the parity of \(tb(L)\)  is determined only by the topology of \(L\) , for example \(tb(S^n)\)  is odd for any Legendrian sphere in \(\mathbb {R}^{2n+1}_{std}\) . To show this, first take the Lagrangian projection of \(L\) , which is an exact Lagrangian immersion in \(\mathbb {R}^{2n}_{std}\) . Notice the parity of \(tb(L)\)  is equal to the \(\bmod \, 2\)  count of self interesections of this Lagrangian immersion, in fact this is an invariant of smooth immersions in \(\mathbb {R}^{2n}\)  up to regular homotopy. Both smooth and Lagrangian immersions satisfy h-principles  [1]}, thus the existence of Lagrangian immersions of a given smooth regular homotopy class is governed by the inclusion map \(\pi _nU_n \rightarrow \pi _nV_{2n,n}\) . For \(n\)  odd this is a map \(\mathbb {Z}\rightarrow \mathbb {Z}_2\) , and (a stable shift of) Lemma  REF  implies this is the zero map except when \(n=1, 3\) .
